From 785478ac23f1ed468e89f79c5886d8c489886077 Mon Sep 17 00:00:00 2001
From: Drew Taylor <drewlynntaylor@gmail.com>
Date: Fri, 21 Jan 2022 00:38:17 -0800
Subject: [PATCH] All finished

---
 hash_practice/exercises.py | 37 +++++++++++++++++++++++++++++++------
 1 file changed, 31 insertions(+), 6 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..a58a4a3 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -2,18 +2,43 @@
 def grouped_anagrams(strings):
     """ This method will return an array of arrays.
         Each subarray will have strings which are anagrams of each other
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(n)
+        Space Complexity: O(n log n)
     """
-    pass
+    letters_hash = {}
+
+    for word in strings: 
+        anagram_sorted = "".join(sorted(word)) 
+        if anagram_sorted in letters_hash: 
+            letters_hash[anagram_sorted].append(word)
+        else: 
+            letters_hash[anagram_sorted] = [word]
+
+    return list(letters_hash.values())
 
 def top_k_frequent_elements(nums, k):
     """ This method will return the k most common elements
         In the case of a tie it will select the first occuring element.
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(n)
+        Space Complexity: O(n)
     """
-    pass
+    highest_k_hash = {}
+    if nums == []:
+        return nums
+
+    for num in nums:
+        if num in highest_k_hash:
+            highest_k_hash[num] += 1
+        else:
+            highest_k_hash[num] = 1
+
+    sorted_k_values = []
+    for i in range(k):
+        highest_freq_value = max(highest_k_hash, key=highest_k_hash.get)
+        sorted_k_values.append(highest_freq_value)
+        highest_k_hash.pop(highest_freq_value)
+
+    return sorted_k_values
 
 
 def valid_sudoku(table):